State the number of valence electrons for an iron(II) ion in an ionic bond.
To determine the number of valence electrons in an iron(II) ion in an ionic bond, we first need to understand the electron configuration of iron (Fe) in its ground state. The atomic number of Fe is 26, so it has 26 electrons in total.
To find the electron configuration, we can refer to the periodic table. Iron is in the 4th period, so its electron configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
To form the iron(II) ion (Fe2+), two electrons are removed from the 4s orbital. This results in the electron configuration:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
In this configuration, the 3d orbital is completely filled with 6 electrons, while the 4s orbital is empty. The valence electrons are located in the outermost occupied shell, which is the 3rd shell in this case. Therefore, the number of valence electrons for an iron(II) ion in an ionic bond is 6.
Remember that valence electrons are the electrons involved in bonding, and in the case of iron(II), they are in the 3d orbital.